Prove that $GL_n(\mathbb{C}) $ is not solvable How should I prove that $GL_n(\mathbb{C}) $, the group of invertible matrices over the complex numbers, is not solvable?
I have no idea how to prove this by supposing that $GL_n(\mathbb{C}) $ is solvable and deducing a contradiction. Does anyone have ideas?
Any hints or advices will help a lot!
 A: Let $K=\Bbb{C}$. The derived group of $GL_n(K)$ is $GL_n(K)^{(1)}=SL_n(K)$, which is perfect for $n\ge 2$. Hence the derived series cannot end with the trivial group, because
$$
GL_n(K)^{(k)}=SL_n(K)\neq 1
$$
for all $k\ge 1$.
A: Soluble groups are closed under taking subgroups. So work in $GL_2(\mathbb{Z})$, which is a subgroup of $GL_n(\mathbb{Z})$, and try and find your favourite non-soluble group as a subgroup. What this subgroup is depends on your background...
My favourite non-soluble subgroup of $GL_2(\mathbb{C})$ is the subgroup generated by the matrices
$$
\begin{align*}
\left(\begin{array}{cc}1&2\\0&1\end{array}\right),\:\:
\left(\begin{array}{cc}1&0\\2&1\end{array}\right)
\end{align*}
$$
which is free of rank two.

An alternative approach is to use finite groups (but this is easiest for $n\geq5$):
The $n\times n$ permutation matrices are those $n\times n$ matrices that have exactly one entry of $1$ in each row and each column and $0$s elsewhere. For each $n$, these form a group isomorphic to $S_n$. Hence, $S_n$ embeds into $GL_n(\mathbb{C})$. For $n\geq5$, $S_n$ is not soluble and so we've found the required non-soluble subgroup.
A: I filled the detail of Dietrich Burde's answer as below:
Note that every commutator of $G=GL_n(\mathbb{C})$ has determinant 1, so $[G,G]\le SL_n(\mathbb{C}).$ Also, by appropriately scaling the elements in $G$(such scaling is justified since $\mathbb{C}$ is algebraically closed.), one can easily check that $[G,G]=[SL_n(\mathbb{C}),SL_n(\mathbb{C})]$. Now it remains to prove that $[SL_n(\mathbb{C}),SL_n(\mathbb{C})]=SL_n(\mathbb{C})$. Since $SL_n(\mathbb{C})$ is generated by the matrices of the form $E_{ij}(\lambda)=I_n+\lambda e_{ij}$, $i\neq j,\:\lambda\in \mathbb{C}$, where $e_{ij}$ is the matrix with the $(i,j)$ entry is 1 and the other entries are zero, it suffices to show that $E_{ij}(\lambda)$ is a commutator of two elements in $SL_n(\mathbb{C})$.
If $n=2$, then 
$$
    \begin{pmatrix}
    1 & \lambda \\
    0 & 1 \\
    \end{pmatrix}=
\begin{pmatrix}
i & 0 \\
0 & -i \\
\end{pmatrix}
\begin{pmatrix}
1 & -\lambda/2 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-i & 0 \\
0 & i \\
\end{pmatrix}
\begin{pmatrix}
1 & \lambda/2 \\
0 & 1 \\
\end{pmatrix}
$$
and
$$
    \begin{pmatrix}
    1 & 0 \\
    \lambda & 1 \\
    \end{pmatrix}=
\begin{pmatrix}
i & 0 \\
0 & -i \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
-\lambda/2 & 1 \\
\end{pmatrix}
\begin{pmatrix}
-i & 0 \\
0 & i \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
\lambda/2 & 1 \\
\end{pmatrix}
$$
and if $n\ge 3$, for each $i\neq j$ we can choose $l$ distinct from both $i$ and $j$, and observe that 
$E_{ij}(\lambda)=E_{il}(\lambda)E_{lj}(1)E_{il}(-\lambda)E_{lj}(-1)$
and we are done. For the calculation of the RHS above, the relation $e_{ij} e_{kl}= \delta_{jk}e_{il}$, where $\delta_{jk}$ is the Kronecker delta, will help.
